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Math Help - General Solution of a Differential Equation

  1. #1
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    General Solution of a Differential Equation

    Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

    Y= (Ce^x)/(x) + (.5x)(e^x).

    Y' + [(1-X)/(X)]Y=e^x.

    However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
    c= -1. How do I arrive at thiis answer?

    Many thanks
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  2. #2
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    Quote Originally Posted by jaijay32 View Post
    Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

    Y= (Ce^x)/(x) + (.5x)(e^x).

    Y' + [(1-X)/(X)]Y=e^x.

    However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
    c= -1. How do I arrive at thiis answer?

    Many thanks
    C=-1, and your particular solution is Y(x) = (-e^x)/(x) + (.5x)(e^x)
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  3. #3
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    Quote Originally Posted by jaijay32 View Post
    Hi. The instructions say "Verify that y is a general solution of the differential equation. Then find the particular solution of the differential equation that satisfies the condition y(1) = -.5e".

    Y= (Ce^x)/(x) + (.5x)(e^x).

    Y' + [(1-X)/(X)]Y=e^x.

    However, when I plug in the point, C always dissapears due to it being associated with a zero that is multiplied. The right answer is that
    c= -1. How do I arrive at thiis answer?

    Many thanks
    If y= (Ce^x)/(x)+ (.5x)e^x then y(1)= (Ce^1)/(1)+ (.5(1))e^1= Ce+ .5e= -.5e. Subtracting .5e from both sides Ce= -e.
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